如何将字节数组转换为十六进制字符串,反之亦然?


当前回答

如果您希望比BitConverter更灵活,但不希望使用那些笨重的90年代风格的显式循环,那么您可以这样做:

String.Join(String.Empty, Array.ConvertAll(bytes, x => x.ToString("X2")));

或者,如果您使用的是.NET 4.0:

String.Concat(Array.ConvertAll(bytes, x => x.ToString("X2")));

(后者来自对原帖子的评论。)

其他回答

具有扩展支持的基本解决方案

public static class Utils
{
    public static byte[] ToBin(this string hex)
    {
        int NumberChars = hex.Length;
        byte[] bytes = new byte[NumberChars / 2];
        for (int i = 0; i < NumberChars; i += 2)
            bytes[i / 2] = Convert.ToByte(hex.Substring(i, 2), 16);
        return bytes;
    }
    public static string ToHex(this byte[] ba)
    {
        return  BitConverter.ToString(ba).Replace("-", "");
    }
}

并像下面那样使用这个类

    byte[] arr1 = new byte[] { 1, 2, 3 };
    string hex1 = arr1.ToHex();
    byte[] arr2 = hex1.ToBin();

这里不想赘述很多答案,但我发现了一个十六进制字符串解析器的相当优化(比公认的好4.5倍)、简单的实现。首先,我的测试输出(第一批是我的实现):

Give me that string:
04c63f7842740c77e545bb0b2ade90b384f119f6ab57b680b7aa575a2f40939f

Time to parse 100,000 times: 50.4192 ms
Result as base64: BMY/eEJ0DHflRbsLKt6Qs4TxGfarV7aAt6pXWi9Ak58=
BitConverter'd: 04-C6-3F-78-42-74-0C-77-E5-45-BB-0B-2A-DE-90-B3-84-F1-19-F6-AB-5
7-B6-80-B7-AA-57-5A-2F-40-93-9F

Accepted answer: (StringToByteArray)
Time to parse 100000 times: 233.1264ms
Result as base64: BMY/eEJ0DHflRbsLKt6Qs4TxGfarV7aAt6pXWi9Ak58=
BitConverter'd: 04-C6-3F-78-42-74-0C-77-E5-45-BB-0B-2A-DE-90-B3-84-F1-19-F6-AB-5
7-B6-80-B7-AA-57-5A-2F-40-93-9F

With Mono's implementation:
Time to parse 100000 times: 777.2544ms
Result as base64: BMY/eEJ0DHflRbsLKt6Qs4TxGfarV7aAt6pXWi9Ak58=
BitConverter'd: 04-C6-3F-78-42-74-0C-77-E5-45-BB-0B-2A-DE-90-B3-84-F1-19-F6-AB-5
7-B6-80-B7-AA-57-5A-2F-40-93-9F

With SoapHexBinary:
Time to parse 100000 times: 845.1456ms
Result as base64: BMY/eEJ0DHflRbsLKt6Qs4TxGfarV7aAt6pXWi9Ak58=
BitConverter'd: 04-C6-3F-78-42-74-0C-77-E5-45-BB-0B-2A-DE-90-B3-84-F1-19-F6-AB-5
7-B6-80-B7-AA-57-5A-2F-40-93-9F

base64和“BitConverter'd”行用于测试正确性。请注意,它们是相等的。

实施:

public static byte[] ToByteArrayFromHex(string hexString)
{
  if (hexString.Length % 2 != 0) throw new ArgumentException("String must have an even length");
  var array = new byte[hexString.Length / 2];
  for (int i = 0; i < hexString.Length; i += 2)
  {
    array[i/2] = ByteFromTwoChars(hexString[i], hexString[i + 1]);
  }
  return array;
}

private static byte ByteFromTwoChars(char p, char p_2)
{
  byte ret;
  if (p <= '9' && p >= '0')
  {
    ret = (byte) ((p - '0') << 4);
  }
  else if (p <= 'f' && p >= 'a')
  {
    ret = (byte) ((p - 'a' + 10) << 4);
  }
  else if (p <= 'F' && p >= 'A')
  {
    ret = (byte) ((p - 'A' + 10) << 4);
  } else throw new ArgumentException("Char is not a hex digit: " + p,"p");

  if (p_2 <= '9' && p_2 >= '0')
  {
    ret |= (byte) ((p_2 - '0'));
  }
  else if (p_2 <= 'f' && p_2 >= 'a')
  {
    ret |= (byte) ((p_2 - 'a' + 10));
  }
  else if (p_2 <= 'F' && p_2 >= 'A')
  {
    ret |= (byte) ((p_2 - 'A' + 10));
  } else throw new ArgumentException("Char is not a hex digit: " + p_2, "p_2");

  return ret;
}

我尝试了一些不安全的东西,并将(显然是冗余的)字符移动到另一个方法来蚕食if序列,但这是最快的。

(我承认这回答了一半的问题。我觉得字符串->字节[]转换不足,而字节[]->字符串角度似乎被很好地覆盖了。因此,这个答案。)

扩展BigInteger方法(Gregory Morse在上面提到过)。我不能评论效率,它使用System.Linq.Reverse(),但它很小而且内置。

        // To hex
        byte[] bytes = System.Text.Encoding.UTF8.GetBytes("Test String!£");
        string hexString = new System.Numerics.BigInteger(bytes.Reverse().ToArray()).ToString("x2");

        // From hex
        byte[] fromHexBytes = System.Numerics.BigInteger.Parse(hexString, System.Globalization.NumberStyles.HexNumber).ToByteArray().Reverse().ToArray();

        // Unit test
        CollectionAssert.AreEqual(bytes, fromHexBytes);
    // a safe version of the lookup solution:       

    public static string ByteArrayToHexViaLookup32Safe(byte[] bytes, bool withZeroX)
    {
        if (bytes.Length == 0)
        {
            return withZeroX ? "0x" : "";
        }

        int length = bytes.Length * 2 + (withZeroX ? 2 : 0);
        StateSmall stateToPass = new StateSmall(bytes, withZeroX);
        return string.Create(length, stateToPass, (chars, state) =>
        {
            int offset0x = 0;
            if (state.WithZeroX)
            {
                chars[0] = '0';
                chars[1] = 'x';
                offset0x += 2;
            }

            Span<uint> charsAsInts = MemoryMarshal.Cast<char, uint>(chars.Slice(offset0x));
            int targetLength = state.Bytes.Length;
            for (int i = 0; i < targetLength; i += 1)
            {
                uint val = Lookup32[state.Bytes[i]];
                charsAsInts[i] = val;
            }
        });
    }

    private struct StateSmall
    {
        public StateSmall(byte[] bytes, bool withZeroX)
        {
            Bytes = bytes;
            WithZeroX = withZeroX;
        }

        public byte[] Bytes;
        public bool WithZeroX;
    }

如果您希望比BitConverter更灵活,但不希望使用那些笨重的90年代风格的显式循环,那么您可以这样做:

String.Join(String.Empty, Array.ConvertAll(bytes, x => x.ToString("X2")));

或者,如果您使用的是.NET 4.0:

String.Concat(Array.ConvertAll(bytes, x => x.ToString("X2")));

(后者来自对原帖子的评论。)